1) McZee climbs up a rope attached to a balloon, pulling the balloon downwards as he ascends. 2) Using the law of conservation of momentum, considering the balloon-monkey system, the balloon's descent can be calculated based on their masses and McZee's ascent distance. 3) The calculation shows that as McZee climbs upwards a distance of L, the balloon will descend a distance of mL/(M+m), where m is McZee's mass and M is the balloon's mass.

1. Chimpoo and McZee - II Our little MacZee …When Chimpoo is not here. He likes to take a little ride with balloons copyrights © youmarks.com

3. Oh come on chose one balloon Mczee wud ya?

5. Thanks balloon. Wait I am coming over to kiss you as a token of appreciation Are you sure? Well than I will also be pulled down anyways towards you if you come towards me. Its ok don’t come. I don’t want to pulled downwards Man this is not cool. I wish I could deflate myself fully to avoid any embarrassment Ok , my friend, I am Acceleration at A m/s2 now upwards. Doing this I will have to pull you downwards which means that as I go up, you come down.

6. Fine. What would be my acceleration downwards McZee? I want to see how much I move downwards for you? I have no idea. Let me Ask his highness Mr. newton Damn it you monkey!! You never rest. Do you? Promise me this is the last time you are doing your stupid climbing.. Ok balloon and monkey , since you both a system together any forces inside the system need not to be considered if I see your overall motion of the system. Now the net force acting on the system is the weights of you two i.e (M+m) g and if I am considering a system the internal forces shall be equal and opposite which is tension force T. So coming to your situation, when you are ascending upwards you are applying a pulling force on rope which in reaction pulls you upwards

7. T mg T Mg Remember these forces are drawn w.ref. t the ground and here in this case if I also consider ground plus you two as a system, there is no external force which is acting on you. So that would help me use the Law of Conservation of Momentum which says “ In absence of any external force acting on the system, the net momentum of the system remains unchanged which is true in deed for your case.Now just check out your momentums, balance them and find out what you are looking for

8. I was initially at rest and not travelling. So my initial momentum was zero. Let us say that I gained a velocity v when I climbed L on the rope , my final momentum w.r.t ground is then mv..What about you balloon? Well, my momentum initially was zero and as you moved upwards, I decended downwards and reached a velocity V when you climbed L upwards w.r.t rope. So my final momentum is MV What do we do now Mr. Newton. Does that means our total momentum should be equated to Zero which means in this case MV-mv = 0 ( taking in to considerations signs of velocities too..Yea this monkey is getting smarter day by day ;) ) Let me check out with the rope now. Rope you might know how much I moved on over you and with what velocity . Any comments where shall we proceed?

9. Yea McZee you moved with a velocity of v+V upwards assuming that M moves down with a velocity V (opposite direction. So L = (v+V)t. And balloon you , you moved down with a velocity V w.r.t ground and the length descended by you is Vt i.e. L = vt + L1 or L1 = vt – L . Put t = L/ (v+V) L1 = v L/ (v+V) – L. Use MV= mv or v /V= M/m L1 = M/m (L) ( M/m +1) – L L1 = mL / (M+m) So, baloon descended mL/ ( M+m) distance downwards the w.r.t ground. I hope that’s clear guys….

10. Well indeed it makes sense, SO I go upwards a distance of L w.r.t balloon and the balloon comes down with a distance of mL/ ( M+m) w.r.t ground .Right.And we meet each other pretty soon as I expected…Love you balloon. I am never holding with this fool again.. And I seriously pity for this rope..Again stuck with this monkey 